Electrostatics and Coulomb's law, Year 5

Late September 2026: classes started a couple of weeks ago and in Year 5 at the art high school (liceo artistico) the teacher has just opened the electrostatics block — charge, Coulomb’s law, followed by electric field and potential. The first test arrives in four to seven weeks, so between the end of October and the first days of November. It is also the year of the maturità (Italy’s school-leaving exam): June 2027 seems very far away, but the syllabus needed to get there starts now, and in physics you recover badly if you start already behind.

At the art high school physics has 2 hours a week, against the 3 of the science high school, and the syllabus is calibrated accordingly: conceptual physics, simple calculations, without the multi-unknown problems seen elsewhere. What follows reflects exactly this slant.

Why it is worth starting from here with the oral exam in mind

In 2026 Physics was one of the four subjects of the oral part of the exam (the colloquio) at the art high school. I don’t yet know whether it will be in 2027 too — the decree with the subjects comes out in January, and stating it today would be making up a fact — but it is worth knowing from now where the thread that opens in these weeks leads: electric charge → electric field → electromagnetic waves → light → colour. It is a link that suits students at an art high school particularly well, because it touches painting and photography directly. The charge you study today is the first link in that chain, even though colour will only arrive in May.

Electric charge: two signs, a minimum quantity

Electric charge exists in two types, which by convention we call positive and negative. Two charges of the same sign repel each other, two of opposite sign attract each other: it is the first thing to make automatic, because in a test a single wrong sign is enough to flip the answer.

Charge is not infinitely divisible: there is a minimum charge, that of the electron (or of the proton, with the opposite sign), $e = 1.6 \times 1 0^{- 19} C$ . Every charge observable in nature is a whole multiple of $e$ — we say that charge is quantised. In addition, charge is conserved: in an isolated system it is neither created nor destroyed, it is transferred from one body to another.

Charging: friction, contact, induction

A neutral body has the same number of positive and negative charges. Charging it means altering that balance, and there are three ways.

By friction: two different materials, rubbed together, exchange electrons — one gives them up, the other gains them. It is the classic case of the balloon rubbed on hair: the balloon pulls electrons from the hair and becomes negatively charged, the hair stays positively charged and for that reason they attract each other (and the balloon).

By contact: a charged body touches a neutral body, and part of the charge passes from the first to the second. The two bodies end up charged with the same sign.

By induction: it is the case that surprises most, because no contact is needed. If you bring the charged balloon (without touching it) close to a thin stream of water coming from the tap, the stream bends towards the balloon — and yet water is neutral. It happens because the field of the charged balloon rearranges the charges already present in the water: it repels them on one side and attracts them on the other, without moving any outwards. The side of the stream closest to the balloon ends up with an excess of charge of opposite sign to the balloon itself, and is attracted. The material stays neutral overall — the charge hasn’t gone anywhere — but it has become polarised: the internal charges have simply rearranged.

Conductors and insulators

In conductors (metals, first of all) the outermost electrons are free to move through the whole material: a charge deposited at one point redistributes itself over the whole surface in an instant. In insulators (glass, plastic, rubber) the electrons stay bound to their atoms and don’t move: that is why, if you rub a plastic ruler at one spot only, the charge stays there, in patches, instead of spreading as it would on a conductor.

Coulomb’s law

The force between two point charges $q_{1}$ and $q_{2}$ , at distance $r$ , has magnitude

F = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}}

with $k = 8.99 \times 1 0^{9} N \cdot m^{2} / C^{2}$ . Three things to hold firm:

it is a vector force: it has a direction (the line joining the two charges) and a sense (attractive if the charges have opposite signs, repulsive if they have the same sign);
the absolute value in the numerator is used to calculate the magnitude of the force, which is always positive: the sign that decides attraction or repulsion must be deduced separately, by looking at the signs of the two charges, not by “bringing it into” the calculation;
the force depends on the inverse square of the distance: it is the same form as universal gravitation, $F = G \frac{m _{1} m _{2}}{r ^{2}}$ .

Comparison with gravity: same form, incomparable strength

The two laws resemble each other in structure ( $1/ r^{2}$ ), but are profoundly different. Gravity is always attractive, Coulomb’s force can be attractive or repulsive depending on the signs. And the strength is not even close: between two electrons, the electric force is about $1 0^{42}$ times stronger than the gravitational force between the same two masses. That is why, at the level of atoms and molecules, gravity is simply negligible: all of chemistry, all of the structure of matter, is held together by electric forces, not gravitational ones.

What happens if the distance changes

This is the question that comes back in every test, in one form or another: if the distance doubles, the force doesn’t halve — it is divided by $2^{2} = 4$ , because the dependence is on the square of the distance. If the distance is halved, the force becomes 4 times larger. If the distance triples, the force is divided by 9. The general rule: if $r$ becomes $n$ times larger, $F$ becomes $n^{2}$ times smaller.

The superposition principle

If several charges act on a charge at the same time, the total force is the vector sum of the forces that each charge would exert on its own, as if the others weren’t there. It is not a sum of numbers: you have to take into account the direction and sense of each contribution.

The dielectric constant of the medium

So far the constant $k$ holds in vacuum (and, in practice, in air too). If the two charges are immersed in another medium — water, for example, or an insulating material — the force between them is weaker, because the medium in turn becomes polarised and partly “screens” the interaction. In the formula a relative dielectric constant $ε_{r} > 1$ of the medium is introduced, which divides the force: the more dielectric the medium, the more the force drops compared with vacuum. It is a detail to keep in mind for now only as a qualitative idea: you always do the calculations in vacuum, unless the text says otherwise explicitly.

A first look at the electric field

Coulomb’s law describes the force between two charges, but it is useful to imagine that every charge generates around itself an electric field, which exists even without a second charge to “feel” it. The field is represented with field lines: they leave positive charges and enter negative ones, denser where the field is stronger. They are the same lines you will draw in a few weeks for the electric potential — it is worth getting your eye used to them from now, because at the art high school visual intuition helps more than the formula.

Three worked exercises

1. Force between two known charges. Two charges $q_{1} = 2 \times 1 0^{- 6} C$ and $q_{2} = 3 \times 1 0^{- 6} C$ are $r = 0.3 m$ apart. Calculate the force between them.

F = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}} = 8.99 \times 1 0^{9} \times \frac{( 2 \times 1 0 ^{- 6} ) ( 3 \times 1 0 ^{- 6} )}{( 0.3 ) ^{2}}

In the numerator: $(2 \times 1 0^{- 6}) (3 \times 1 0^{- 6}) = 6 \times 1 0^{- 12}$ . In the denominator: $(0.3)^{2} = 0.09$ . So

F = 8.99 \times 1 0^{9} \times \frac{6 \times 1 0 ^{- 12}}{0.09} = 8.99 \times 1 0^{9} \times 6.67 \times 1 0^{- 11} \approx 0.6 N

The two charges are both positive: the force is repulsive.

2. The distance doubles. Same charges as the previous exercise, but now $r = 0.6 m$ (double the previous). By how much does the force change, without redoing the whole calculation?

The distance has doubled, so the force is divided by $2^{2} = 4$ :

F^{'} = \frac{F}{4} = \frac{0.6}{4} = 0.15 N

3. Three charges in a line. On a line, in order, there are $q_{1} = + 4 \times 1 0^{- 6} C$ , then at a distance of $0.2 m$ $q_{2} = - 2 \times 1 0^{- 6} C$ (the middle charge), then at another $0.2 m$ $q_{3} = + 4 \times 1 0^{- 6} C$ . Find the resultant force on $q_{2}$ .

The force of $q_{1}$ on $q_{2}$ :

F_{1} = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}} = 8.99 \times 1 0^{9} \times \frac{( 4 \times 1 0 ^{- 6} ) ( 2 \times 1 0 ^{- 6} )}{( 0.2 ) ^{2}} = 8.99 \times 1 0^{9} \times \frac{8 \times 1 0 ^{- 12}}{0.04} \approx 1.8 N

$q_{1}$ is positive and $q_{2}$ negative: the force is attractive, so $q_{2}$ is pulled towards $q_{1}$ (to the left, if $q_{1}$ is on the left).

The force of $q_{3}$ on $q_{2}$ : same distance, same charges in magnitude, so $F_{3} = 1.8 N$ too, but directed towards $q_{3}$ (to the right), because $q_{3}$ is also positive and attracts the negative $q_{2}$ from the other side.

The two forces have the same magnitude and opposite sense: they cancel. The resultant force on $q_{2}$ is zero. It is a case where the calculation is deliberately simple, to show that superposition is a sum of vectors and not of numbers: adding $1.8 + 1.8 = 3.6 N$ while ignoring the directions would be the classic mistake.

Typical mistakes

Mistake	How you notice it
Putting the signs inside the absolute value and getting a negative force	The magnitude of a force can’t be negative: if you get a negative number, you have mishandled the signs
Getting the power of ten wrong in the product of the charges	Rewrite each charge as $a \times 1 0^{n}$ and multiply the $a$ first, then add the exponents: don’t trust your eye
Using centimetres instead of metres for the distance	The constant $k$ is defined for $r$ in metres: always convert before substituting into the formula
Adding forces as numbers instead of as vectors	Always draw an arrow for each force before adding: if two forces have opposite senses, they subtract, they don’t add
Confusing charge and current	Charge ( $C$ ) is a quantity; current ( $A = C / s$ ) is charge flowing over time — they are two different quantities, with different symbols

A note on the linked practice test

The electrostatics practice test on the site is catalogued under Year 4, because at the science high school this topic is done in Year 4. At the art high school it arrives in Year 5, but the electrostatics exercises are identical — you can use it without any problem to practise on charge and Coulomb’s law. The part of that practice test devoted to circuits you will meet later, in January.

If you also want to review the rest of the physics syllabus, you’ll find everything at /materie/fisica/. And if you come from the science track and are looking for the same topic with a different slant — more calculations, more unknowns — the twin post is Electrostatics and circuits: preparing for the Year 4 test, written for those who study it a year earlier and with an extra hour a week. If instead you have already been through the first physics test in Year 3 at art high school, the post on measurements and significant figures covers the topic that preceded it; and for maths on the same track there is numbers, fractions and powers in Year 1.

If you want a hand setting up this block before the end-of-October test, the first introductory call is free and lasts half an hour.

Electrostatics and Coulomb's law in Year 5 of high school